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Let $V$ be the vector space of $2 \times 2$ matrices over $\mathbb F$. Let $W_1$ be the set of matrices of the form $\begin{bmatrix} x &−x \\ y & z \end{bmatrix}$ and let $W_2$ be the set of matrices of the form $\begin{bmatrix} a &b \\ −a & c \end{bmatrix}$. What is the dimension of $W_1+W_2$?

I found that the basis of $W_1$ and the basis of $W_2$ have size $3$, so $\dim W_1=\dim W_2=3$. I also found that the basis of $W_1 \cap W_2$ has size $2$, so $\dim(W_1 \cap W_2)=2$.

From this informatinon, how can I find the dimension of $W_1+W_2$ (without using the formula for the dimension of a sum)?

All I know so far is that $W_1+W_2$ is the smallest subspace that contains $W_1$ and $W_2$, and is contained in $V$. Since $\dim W_1=\dim W_2=3$ and $\dim V=4$, we must have $\dim(W_1+W_2)$ to be $3$ or $4$.

How do I know which one is it?

Source: Linear Algebra by Hoffman and Kunze - Exercise 7 of Section 2.3

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  • $\begingroup$ If $\beta_1$ is a basis for $W_1$ and $\beta_2$ is a basis for $W_2$, then it should be easy for you to verify that $\beta_1 \cup \beta_2$ will span $W_1+W_2$. So, to find the dimension of this space, all you need to do is find the number of linearly independent vectors in $\beta_1 \cup \beta_2$. In this particular example, it is not hard to find an explicit basis for $W_1 + W_2$ (after finding the basis, you'll see the dimension is $4$). $\endgroup$ – peek-a-boo Jul 9 '19 at 0:56
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The dimension is $4$, that is, $W_1+W_2$ consists of all the $2\times 2$ matrices. You can check directly that every matrix can be written as a sum of a matrix from $W_1$ and a matrix from $W_2$, for instance: $$\begin{bmatrix} x &y \\ z & t \end{bmatrix}=\begin{bmatrix} x &-x \\ z & t \end{bmatrix}+ \begin{bmatrix} 0 & x+y \\ 0 & 0 \end{bmatrix}$$

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Alternatively, still without using the inclusion-exclusion dimension formula, note that $$W_1 = W_1 + \{0\} \subseteq W_1 + W_2.$$ If $\dim(W_1 + W_2) = 3 = \dim(W_1)$, then $W_1$ is a subspace of $W_1 + W_2$, but with the same dimension. This implies that $W_1 = W_1 + W_2$. A similar argument also shows that $W_2 = W_1 + W_2$. So, under this assumption, we have $$W_1 = W_2.$$ This is very easy to disprove!

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